A branching particle system approximation for solving partially observed stochastic optimal control problems via stochastic maximum principle

Abstract

This paper develops an efficient numerical algorithm for solving a class of partially observed stochastic optimal control problems with correlated noises. The main contribution of this paper is threefold: first, we introduce a relaxed system and assume the Roxin condition (convexity requirement) on coefficients. Then, an optimal relaxed system provides an optimal admissible control in a broader sense, and a relaxed control turns out to be a usual admissible control. Second, we transform the optimal control problem into an optimization problem for a convex functional by employing a projection operator. A stochastic gradient descent approach is then proposed and its convergence properties are demonstrated. Last but not least, we present a branching particle system (branching particle filter) to approximate the optimal filter. Due to the random nature of the coefficients in the Zakai equation, neither the dual approach nor the mild solution approach can be used. We devise a novel method for establishing the convergence of the branching particle system approximation, as well as its rate of convergence. This branching-type particle filter algorithm allows us to tackle non-Markovian environments. The major body of this paper concludes with a numerical case study.

Appendices

Appendix A: Proof of Theorem 6

We will use two steps to finish the proof. In this appendix, $\mathbb {E}$ represents $\mathbb {E}^{\mathbb {Q}}$.

We begin by proving it for the special case that $\tau$ takes values only in a countable set of times $\{t_1,t_2,\ldots \}$. We need to show that $\mathbb {E}[\pi _{\tau }(\phi )\textbf{1}_{A}]=\mathbb {E}[\phi (\mathcal {X}_{\tau })\textbf{1}_{A}]$ for every $A\in \mathcal {F}_{\tau }^{Y}$. Note that $A=\bigcup _{i\ge 1} A\cap \{\tau =t_i\} \triangleq \bigcup _{i\ge 1} A_i$ implies that

$$\begin{aligned} \begin{aligned}&\mathbb {E}[\pi _{\tau }(\phi )\textbf{1}_{A}] = \sum _{i=1}^{\infty } \mathbb {E}\left[ \pi _{\tau }(\phi )\textbf{1}_{A\cap \{t=t_i\}} \right] , \\&\mathbb {E}[\phi (\mathcal {X}_{\tau })\textbf{1}_{A}] = \sum _{i=1}^{\infty } \mathbb {E}\left[ \phi (\mathcal {X}_{\tau })\textbf{1}_{A\cap \{t=t_i\}} \right] . \end{aligned} \end{aligned}$$

Then it follows from $A_i\in \mathcal {F}_{t_i}^{Y}$ that

$$\begin{aligned} \begin{aligned} \mathbb {E}[\pi _{\tau }(\phi )\textbf{1}_{A_i}] = \mathbb {E}[\pi _{t_i}(\phi )\textbf{1}_{A_i}] = \mathbb {E}[\phi (\mathcal {X}_{t_i})\textbf{1}_{A_i}] = \mathbb {E}[\phi (\mathcal {X}_{\tau })\textbf{1}_{A_i}]. \end{aligned} \end{aligned}$$

Moreover,

$$\begin{aligned} \begin{aligned} \{ \pi _{\tau }(\phi )\in B \} = \bigcup _{i\ge 0} \{ \pi _{\tau }(\phi )\in B~\text {and}~\tau =t_i \} = \bigcup _{i\ge 0} \{ \pi _{t_i}(\phi )\in B \} \cap \{ \tau =t_i \} \end{aligned} \end{aligned}$$

for every Borel set B; hence $\{ \pi _{\tau }(\phi )\in B \} \cap \{ \tau =t_j \} \in \mathcal {F}_{t_j}^{Y}\subset \mathcal {F}_{t_i}^{Y}$ for every $j\le i$. Consequently, $\pi _{\tau }(\phi )=\bigcup _{j\le i} \{ \pi _{\tau }(\phi )\in B \} \cap \{ \tau =t_j \}$ is $\mathcal {F}_{\tau }^{Y}$-measurable. This completes the proof in discrete time case.

We proceed to the proof in continuous time case. Define the stopping times $\tau _n=([2^n\tau ]+1)/2^n$. Then $\tau _n\downarrow \tau$, and each $\tau _n$ takes a countable number of values. It follows immediately that

$$\begin{aligned} \begin{aligned} \pi _{\tau _n}(\phi )=\mathbb {E}[\phi (\mathcal {X}_{\tau _n})|\mathcal {F}_{\tau _n}^{Y}] \end{aligned} \end{aligned}$$

(48)

for every n. By the continuity of $\pi _{t}(\phi )$, we take the limit in the left-hand side of (48) and obtain

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty } \pi _{\tau _n}(\phi ) = \pi _{\tau }(\phi ). \end{aligned} \end{aligned}$$

It remains to tackle the right-hand side of (48). We claim that $\mathcal {F}_{\tau _{n+1}}^{Y}\subset \mathcal {F}_{\tau _{n}}^{Y}$. Since $\tau _{n+1}\le \tau _n$ and $\tau _n$ is an $\mathcal {F}_{t}^{Y}$-stopping time, we have

$$\begin{aligned} \begin{aligned} C\cap \{ \tau _n\le t \} = C \cap \{ \tau _{n+1}\le t \} \cap \{ \tau _n\le t \} \in \mathcal {F}_{t}^{Y} \end{aligned} \end{aligned}$$

for every $C\in \mathcal {F}_{\tau _{n+1}}^{Y}$. Therefore, the claim follows from the definition of $\mathcal {F}_{\tau _n}^{Y}$. According to Hunt’s Lemma, we draw a nontrivial conclusion

$$\begin{aligned} \begin{aligned} \mathbb {E}[\phi (\mathcal {X}_{\tau _n})|\mathcal {F}_{\tau _n}^{Y}] \rightarrow \mathbb {E}[\phi (\mathcal {X}_{\tau })|\mathcal {G}], \quad \text {in } L^2, \text { as } n\rightarrow \infty , \end{aligned} \end{aligned}$$

where $\mathcal {G}=\bigcap _{n\ge 1}\mathcal {F}_{\tau _n}^{Y}$. We now have to show that $\mathbb {E}[\phi (\mathcal {X}_{\tau })|\mathcal {G}]=\mathbb {E}[\phi (\mathcal {X}_{\tau })|\mathcal {F}_{\tau }^{Y}]$. It clearly that $\mathcal {F}_{\tau }^{Y}\subset \mathcal {G}$. Hence it remains to prove $\mathcal {F}_{\tau }^{Y}\supset \mathcal {G}$. Note that $\pi _{\tau }(\phi )=\mathbb {E}[\phi (\mathcal {X}_{\tau })|\mathcal {G}]$. We desire to verify that $\pi _{\tau }(\phi )$ is $\mathcal {F}_{\tau }^{Y}$-measurable. Define the stopping times $\sigma _n=\tau _n-2^{-n}$. Then $\sigma _n\le \tau$, and $\sigma _n\uparrow \tau$ as $n\rightarrow \infty$. But $\pi _{\sigma _n}(\phi )$ is $\mathcal {F}_{\sigma _n}^{Y}$-measurable, so it is $\mathcal {F}_{\tau }^{Y}$-measurable for every n. Thus, $\pi _{\tau }(\phi )=\lim _{n\rightarrow \infty }\pi _{\sigma _n}(\phi )$ must be $\mathcal {F}_{\tau }^{Y}$-measurable. We conclude that $\mathbb {E}[\mathbb {E}[\phi (\mathcal {X}_{\tau })|\mathcal {G}]|\mathcal {F}_{\tau }^{Y}] =\mathbb {E}[\phi (\mathcal {X}_{\tau })|\mathcal {G}]$, i.e., $\mathcal {F}_{\tau }^{Y}\supset \mathcal {G}$. Then the proof is complete. $\square$

Appendix B: Proof of Lemma 5

Let us define the process $q_{m}^{n}(t)$ as a unique $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};V_m)$ valued solution to the system

$$\begin{aligned} \begin{aligned}&(q_{m}^{n}(t),e_{\ell })_{H^0} \\&\quad = (q_{m}^{n}(k\delta ),e_{\ell })_{H^0} + \int _{k\delta }^{t} {_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(r),e_{\ell } \right\rangle _{H^1} \textrm{d}r + \int _{k\delta }^{t} \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0} \textrm{d}Y_r \\&\qquad + \frac{1}{n} \sum _{j=1}^{n} \int _{k\delta }^{t} \xi _{k\delta }^{n} M_j^n(r) \left[ R_{r}^{1}(e_{\ell })\textrm{d}r + R_{r}^{2}(e_{\ell })\textrm{d}Y_r + R_{r}^{3}(e_{\ell }) \textrm{d}W_r^{j} \right] . \end{aligned} \end{aligned}$$

(49)

The well-posedness of the system (49) does not follow directly from the standard theory for SODEs, since the coefficients of that equation need not be Lipschitz. However, various conclusions, for example, Theorem 3.21 in Pardoux and Răşcanu [39], allow us to deal with the current predicament. Assuming that the system (49) is well-posed for the time being, the next step is to build a uniform bound on the family $\{q_{m}^{n}(t)\}_{m\ge 1}$.

In this appendix, $\mathbb {E}$ stands for $\mathbb {E}^{\mathbb {P}}$. We now establish a sufficiently strong a priori bound

$$\begin{aligned} \begin{aligned} \sup _{m\ge 1} \mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \Vert q_{m}^{n}(t) \Vert _{H^0}^{2} + \int _{k\delta }^{(k+1)\delta } \Vert q_{m}^{n}(r) \Vert _{H^1}^{2} \textrm{d}r \right] < \infty . \end{aligned} \end{aligned}$$

(50)

With the use of (49) and Itô’s formula, we can get the identity for all $1\le \ell \le m$

$$\begin{aligned}&(q_{m}^{n}(t),e_{\ell })_{H^0}^2 \\&\quad = (q_{m}^{n}(k\delta ),e_{\ell })_{H^0}^2 + \int _{k\delta }^{t} \left[ 2\left( q_{m}^{n}(r),e_{\ell }\right) _{H^0} {_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(r),e_{\ell } \right\rangle _{H^1} + \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0}^{2} \right] \textrm{d}r \\&\qquad + \int _{k\delta }^{t} 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{1}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \int _{k\delta }^{t} \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \int _{k\delta }^{t} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell }) \right) ^2 \textrm{d}r \\&\qquad + \int _{k\delta }^{t} \left[ 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0} + 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \right. \\&\qquad \left. \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \right] \textrm{d}Y_r \\&\qquad + \int _{k\delta }^{t} 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell })\right) \textrm{d}W_{r}^{j}. \end{aligned}$$

By adding the values from $\ell =1$ to $\ell =m$, we obtain the following result

$$\begin{aligned} \begin{aligned}&\left\| q_{m}^{n}(t) \right\| _{H^0}^2 \\&\quad = \sum _{\ell =1}^{m} (q_{m}^{n}(k\delta ),e_{\ell })_{H^0}^2 + \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(r),q_{m}^{n}(r) \right\rangle _{H^1} + \left\| \mathcal {B} q_{m}^{n}(r) \right\| _{H^0}^{2} \right] \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \int _{k\delta }^{t} 2(q_{m}^{n}(r),e_{\ell }) \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{1}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \int _{k\delta }^{t} \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \int _{k\delta }^{t} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell }) \right) ^2 \textrm{d}r \\&\qquad + \int _{k\delta }^{t} \left[ 2\left( \mathcal {B} q_{m}^{n}(r),q_{m}^{n}(r) \right) _{H^0} + 2\sum _{\ell =1}^{m} \left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \right] \textrm{d}Y_r \\&\qquad + \sum _{\ell =1}^{m} \int _{k\delta }^{t} 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell })\right) \textrm{d}W_{r}^{j}. \end{aligned} \end{aligned}$$

(51)

In order to take the expectation and eliminate the stochastic integrals, we first need to create a sequence of stopping times

$$\begin{aligned} \begin{aligned} \tau _R:= \inf \left\{ t\in [k\delta ,(k+1)\delta ): \Vert q_{m}^{n}(t) \Vert _{H^0}^{2} \vee \int _{k\delta }^{t} \Vert q_{m}^{n}(r) \Vert _{H^1}^{2} \textrm{d}r \ge R \right\} . \end{aligned} \end{aligned}$$

We observe that $\tau _R\rightarrow \infty$, $\mathbb {P}$-a.s., as $R\rightarrow \infty$. Consider the stopped process $q_{m}^{n}(t\wedge \tau _R)$, for which we have the bound

$$\begin{aligned} \begin{aligned}&\mathbb {E} \Vert q_{m}^{n}(t\wedge \tau _R) \Vert _{H^0}^2 \\&\quad \le \mathbb {E} \Vert q_{m}^{n}(k\delta ) \Vert _{H^0}^2 + \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \left[ 2{_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(r),q_{m}^{n}(r) \right\rangle _{H^1} + \left\| \mathcal {B} q_{m}^{n}(r) \right\| _{H^0}^{2} \right] \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{1}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell }) \right) ^2 \textrm{d}r. \end{aligned} \end{aligned}$$

It should be highlighted that $n\gg m$ (for instance, take ${\widetilde{n}}=n\cdot m$), then it is straightforward to demonstrate that

$$\begin{aligned} \begin{aligned}&\sum _{\ell =1}^{m} \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} 2\left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{1}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \left( \mathcal {B} q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \textrm{d}r \\&\qquad + \sum _{\ell =1}^{m} \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell }) \right) ^2 \textrm{d}r \\&\quad \lesssim \frac{1}{n} + \mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \Vert q_{m}^{n}(r) \Vert _{H^0}^{2} \textrm{d}r. \end{aligned} \end{aligned}$$

Combine the resulting inequality with the coercive condition (see (9)) to yield

$$\begin{aligned} \begin{aligned}&\mathbb {E}\left[ \Vert q_{m}^{n}(t\wedge \tau _R) \Vert _{H^0}^2 + C_1 \int _{k\delta }^{t\wedge \tau _R} \Vert q_{m}^{n}(r) \Vert _{H^1}^{2} \textrm{d}r \right] \\&\quad \lesssim \mathbb {E} \Vert q_{m}^{n}(k\delta ) \Vert _{H^0}^2 + \frac{1}{n} + (1+C)\mathbb {E} \int _{k\delta }^{t\wedge \tau _R} \Vert q_{m}^{n}(r) \Vert _{H^0}^2 \textrm{d}r. \end{aligned} \end{aligned}$$

Thus, we infer that

$$\begin{aligned} \begin{aligned} \sup _{m\ge 1}\sup _{t\in [k\delta ,(k+1)\delta )} \mathbb {E}\left[ \Vert q_{m}^{n}(t) \Vert _{H^0}^{2} + C_1 \int _{k\delta }^{t} \Vert q_{m}^{n}(t) \Vert _{H^1}^{2} \textrm{d}t \right] < \infty . \end{aligned} \end{aligned}$$

Obviously, we merely need to exchange the supremum over $[k\delta ,(k+1)\delta )$ with the expectation. It follows from Buckholder–Davis–Gundy’s inequality as well as Hölder’s inequality that

$$\begin{aligned} \begin{aligned}&\mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \left| \int _{k\delta }^{t} \left( \mathcal {B}q_{m}^{n}(r),q_{m}^{n}(r) \right) _{H^0} \textrm{d}Y_r \right| \right] \\&\quad \le C \mathbb {E} \left[ \int _{k\delta }^{(k+1)\delta } \left( \mathcal {B}q_{m}^{n}(r),q_{m}^{n}(r) \right) _{H^0}^{2} \textrm{d}r \right] ^{1/2} \\&\quad \le C \mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \Vert q_{m}^{n}(t) \Vert _{H^0} \sqrt{\int _{k\delta }^{(k+1)\delta } |\mathcal {B}q_{m}^{n}(r)|^2 \textrm{d}r} \right] \\&\quad \le \frac{1}{2} \mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \Vert q_{m}^{n}(t) \Vert _{H^0}^{2} \right] + \frac{C^2}{2} \mathbb {E} \int _{k\delta }^{(k+1)\delta } |\mathcal {B}q_{m}^{n}(r)|^2 \textrm{d}r. \end{aligned} \end{aligned}$$

Similarly, one deduces that

$$\begin{aligned} \begin{aligned}&\mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \left| \sum _{\ell =1}^{m} \int _{k\delta }^{t} 2 \left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) \textrm{d}Y_r \right| \right] \\&\quad \le C \mathbb {E}\left[ \sum _{\ell =1}^{m} \int _{k\delta }^{t} \left( q_{m}^{n}(r),e_{\ell } \right) _{H^0}^{2} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{2}(e_{\ell }) \right) ^2 \textrm{d}r \right] \\&\quad \lesssim \frac{1}{n} \mathbb {E} \int _{k\delta }^{t} \Vert q_{m}^{n}(r) \Vert _{H^0}^{2} \textrm{d}r, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \left| \sum _{\ell =1}^{m} \int _{k\delta }^{t} 2 \left( q_{m}^{n}(r),e_{\ell } \right) _{H^0} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell }) \right) \textrm{d}W_{r}^{j} \right| \right] \\&\quad \le C \mathbb {E}\left[ \sum _{\ell =1}^{m} \int _{k\delta }^{t} \left( q_{m}^{n}(r),e_{\ell } \right) _{H^0}^{2} \left( \frac{1}{n} \sum _{j=1}^{n} \xi _{k\delta }^{n} M_j^n(r)R_{r}^{3}(e_{\ell }) \right) ^2 \textrm{d}r \right] \\&\quad \lesssim \frac{1}{n} \mathbb {E} \int _{k\delta }^{t} \Vert q_{m}^{n}(r) \Vert _{H^0}^{2} \textrm{d}r. \end{aligned} \end{aligned}$$

If we take a supremum over $t\in [k\delta ,(k+1)\delta )$, prior to taking the expectation in (51), we therefore gain

$$\begin{aligned} \begin{aligned}&\mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \Vert q_{m}^{n}(t) \Vert _{H^0}^{2} \right] \\&\quad \lesssim \mathbb {E} \Vert q_{m}^{n}(k\delta ) \Vert _{H^0}^{2} \\&\qquad + \mathbb {E} \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(r),q_{m}^{n}(r) \right\rangle _{H^1} + (1+C^2)\left\| \mathcal {B} q_{m}^{n}(r) \right\| _{H^0}^{2} \right] \textrm{d}r \\&\qquad + \frac{1}{n} \mathbb {E} \int _{k\delta }^{t} \Vert q_{m}^{n}(r) \Vert _{H^0}^{2} \textrm{d}r. \end{aligned} \end{aligned}$$

Employing the coercive condition, we have

$$\begin{aligned} \begin{aligned} \sup _{m\ge 1} \mathbb {E}\left[ \sup _{t\in [k\delta ,(k+1)\delta )} \Vert q_{m}^{n}(t) \Vert _{H^0}^{2} \right] \lesssim \mathbb {E} \Vert q_{m}^{n}(k\delta ) \Vert _{H^0}^{2} + \sup _{m\ge 1} \mathbb {E} \int _{k\delta }^{t} \Vert q_{m}^{n}(r) \Vert _{H^0}^{0} \textrm{d}r. \end{aligned} \end{aligned}$$

Hence, we ultimately establish the a priori bound (50).

According to (50) and the linear growth of $\mathcal {A}^v$ (see (11)), one obtains that

• (i) $\{q_{m}^{n}(t)\}_{m\ge 1}$ is uniformly bounded in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^1)$;

• (ii) $\{\mathcal {A}^v q_{m}^{n}(t)\}_{m\ge 1}$ is uniformly bounded in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^{-1})$;

• (iii) $\{\mathcal {B} q_{m}^{n}(t)\}_{m\ge 1}$ is uniformly bounded in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^0)$.

Thus, the Banach–Alaoglu theorem implies that there are some weakly convergent subsequences. We do not relabel the indices with respect to m of the following convergent subsequences such that

• (i) $q_{m}^{n}(t)\rightharpoonup q^n(t)$ in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^1)$;

• (ii) $\mathcal {A}^v q_{m}^{n}(t)\rightharpoonup \zeta (t)$ in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^{-1})$;

• (iii) $\mathcal {B} q_{m}^{n}(t)\rightharpoonup \eta (t)$ in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^0)$.

It merely remains to verify that $\zeta (t)=\mathcal {A}^v q^n(t)$ and $\eta (t)=\mathcal {B} q^n(t)$. Additionally, we have the fact that $q_{m}^{n}(t)\overset{*}{\rightharpoonup }q^n(t)$ in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};L^{\infty })$. By the dominated convergence theorem, we may pass to the limit as $m\rightarrow \infty$ in (51) $(n\rightarrow \infty ~\text {as well})$

$$\begin{aligned} \begin{aligned} \mathbb {E}\left[ \Vert q_{}^{n}((k+1)\delta ) \Vert _{H^0}^{2} - \Vert q_{}^{n}(k\delta ) \Vert _{H^0}^{2} \right]&= \mathbb {E} \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \zeta (r),q_{}^{n}(r) \right\rangle _{H^1} + \left\| \eta (r) \right\| _{H^0}^{2} \right] \textrm{d}r. \end{aligned} \end{aligned}$$

Since the map $\varrho \rightarrow \mathbb {E}\Vert \varrho \Vert _{H^0}^{2}$ is convex, we get that

$$\begin{aligned} \begin{aligned} \mathbb {E}\left[ \Vert q_{}^{n}((k+1)\delta ) \Vert _{H^0}^{2} - \Vert q_{}^{n}(k\delta ) \Vert _{H^0}^{2} \right]&\le \liminf _{m\rightarrow \infty } \mathbb {E}\left[ \Vert q_{m}^{n}((k+1)\delta ) \Vert _{H^0}^{2} - \Vert q_{m}^{n}(k\delta ) \Vert _{H^0}^{2} \right] . \end{aligned} \end{aligned}$$

Then we have the inequality

$$\begin{aligned} \begin{aligned}&\mathbb {E} \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \zeta (r),q_{}^{n}(r) \right\rangle _{H^1} + \left\| \eta (r) \right\| _{H^0}^{2} \right] \textrm{d}r \\&\quad \le \liminf _{m\rightarrow \infty } \mathbb {E} \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(r),q_{m}^{n}(r) \right\rangle _{H^1} + \left\| \mathcal {B} q_{m}^{n}(r) \right\| _{H^0}^{2} \right] \textrm{d}r. \end{aligned} \end{aligned}$$

(52)

From the monotonicity condition (see (10)) with $\lambda =0$, for all $p(t)\in L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^1)$ and $m\ge 1$, we derive that

$$\begin{aligned} \begin{aligned} \mathbb {E} \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \mathcal {A}^v q_{m}^{n}(t) - \mathcal {A}^v p(t),q_{m}^{n}(t) - p(t) \right\rangle _{H^1} + \left\| \mathcal {B}q_{m}^{n}(t) - \mathcal {B}p(t) \right\| _{H^0}^{2} \right] \textrm{d}r \le 0. \end{aligned}\nonumber \\ \end{aligned}$$

(53)

With the assistance of (52) and (53), and proceeding in a similar manner to the limit, we have

$$\begin{aligned} \begin{aligned} \mathbb {E} \int _{k\delta }^{t} \left[ 2{_{H^{-1}}}\left\langle \zeta (t) - \mathcal {A}^v p(t),q_{}^{n}(t) - p(t) \right\rangle _{H^1} + \left\| \eta (t) - \mathcal {B}p(t) \right\| _{H^0}^{2} \right] \textrm{d}r \le 0. \end{aligned}\nonumber \\ \end{aligned}$$

(54)

If we set $q_{}^{n}(t)=p(t)$ in (54) we immediately obtain $\eta (t)=\mathcal {B}q_{}^{n}(t)$. To show that $\zeta (t)=\mathcal {A}^v q_{}^{n}(t)$, let $p(t)=q_{}^{n}(t) - \theta w(t)$ for $\theta >0$ and $w(t)\in L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^1)$. Diving both side by $\theta$ we see that

$$\begin{aligned} \begin{aligned} \mathbb {E} \int _{k\delta }^{(k+1)\delta } {_{H^{-1}}}\left\langle \zeta (r) - \mathcal {A}^v(q_{}^{n}(r)-\theta w(r)),w(r) \right\rangle _{H^1} \textrm{d}r \le 0. \end{aligned} \end{aligned}$$

Letting $\theta \rightarrow 0$ and utilizing the weak continuity of $\mathcal {A}^v$ (see (12)), we deduce that

$$\begin{aligned} \begin{aligned} \mathbb {E} \int _{k\delta }^{(k+1)\delta } {_{H^{-1}}}\left\langle \zeta (r) - \mathcal {A}^v q_{}^{n}(r),w(r) \right\rangle _{H^1} \textrm{d}r \le 0, \quad \forall ~w(t)\in L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^1). \end{aligned} \end{aligned}$$

It follows unambiguously that $\zeta (t)=\mathcal {A}^v q^n(t)$ in $L^2([0,T]\times \varOmega ,\mathcal {F}_{t}^{Y};H^{-1})$.

The proof of the theorem has been done. $\square$